\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.05................................0.05\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(n_{CuO}=\dfrac{6}{80}=0.075\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(1............1\)

\(0.075......0.05\)

Chất khử : H₂ . Chất OXH : CuO 

\(LTL:\dfrac{0.075}{1}>\dfrac{0.05}{1}\Rightarrow CuOdư\)

\(m_{CuO\left(dư\right)}=\left(0.075-0.05\right)\cdot64=1.6\left(g\right)\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
b)

Theo PTHH : 

$n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$

$V_{H_2} = 0,1.22,4 = 2,24(lít)$

c)

$n_{CuO} = \dfrac{8}{80} = 0,1(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy : $\dfrac{n_{CuO}}{1} = \dfrac{n_{H_2}}{1}$ nên CuO phản ứng hết

$n_{Cu} = n_{H_2} = 0,1(mol)$
$m_{Cu} = 0,1.64 = 6,4(gam)$

\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\) 

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\) 

                1        2               1           1

             0,05      0,1          0,05       0,05

a) \(V_{H_2}=n.24,79=0,05.24,79=1,2395\left(l\right)\) 

\(m_{ZnCl_2}=n.M=0,05.\left(65+35,5.2\right)=6,8\left(g\right)\) 

b) \(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 

Ta cos tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\Rightarrow\) CuO dư.

Theo ptr, ta có: \(n_{Cu}=n_{H_2}=0,05mol\\ \Rightarrow m_{Cu}=n.M=0,05.64=3,2\left(g\right).\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

THeo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,05\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,05.24,79=1,2395\left(l\right)\)

\(m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)

b, Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,05\left(mol\right)\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)

Theo gt ta có: $n_{Zn}=0,1(mol);n_{CuO}=0,25(mol)$

a, $Zn+2HCl\rightarrow ZnCl_2+H_2$

$CuO+H_2\rightarrow Cu+H_2O$

b, Ta có: $n_{ZnCl_2}=0,1(mol)\Rightarrow m_{ZnCl_2}=13,6(g)$

b, Ta có: $n_{H_2}=0,1(mol)$ 

Sau phản ứng chất còn dư là CuO dư 0,15 mol

$\Rightarrow m_{CuO/du}=12(g)$

a)

$Zn + 2HCl \to ZnCl_2 + H_2$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

b)

n ZnCl2 = n Zn = 6,5/65 = 0,1(mol)

=> m ZnCl2 = 0,2.136 = 13,6(gam)

c) n H2 = n Zn = 0,1 mol

CuO + H2 --to--> Cu + H2O

n CuO = 20/80 = 0,25 > n H2 = 0,1 nên CuO dư

n CuO pư = n H2 = 0,1 mol

=> m CuO dư = 20 - 0,1.80 = 12(gam)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

b, Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

c, \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\), ta được CuO dư.

Theo PT: \(n_{CuO\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)

Số mol của 5,6g Fe:

\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

a,\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

b,thể tích của 0,1 mol H2:

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

c,\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\) 

khối lượng của 0,1 mol Cu:

\(m_{Cu}=n.M=0,1.64=6,4\left(g\right)\)

a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)

b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)

Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

\(m_{CuO}=0,5.80=40\left(g\right)\)

nMg = 4.8/24 = 0.2 (mol) 

Mg + 2HCl => MgCl2 + H2 

0.2.................................0.2

CuO + H2 -to-> Cu + H2O 

...........0.2..........0.2

mCu = 0.2*64 = 12.8 (g) 

a) PTHH: Mg +  2HCl -> MgCl2 + H2

0,2____________0,4___0,2___0,2(mol)

CuO + H2 -to-> Cu + H2O

0,2___0,2____0,2(mol)

b) =>mCu=0,2.64=12,8(g)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)